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Any probability measure $\mu_1$ absolutely continuous with respect to $\mu_1$ can be written as a Gibbs measure if you allow $G$ to take values $\pm \infty$. If the density is bounded above and below, $G$ will be bounded. So you're basically asking about how to sample from a probability measure. This is a big field of study. Markov chain Monte Carlo (MCMC)...

10

Maybe I'm missing something, but it seems to me you're overthinking this. Since $$\int_{0}^{1} d\theta (1-\theta ) \theta^{n} = \frac{1}{n+1} -\frac{1}{n+2} = \frac{n!}{(n+2)!}$$ the third term in your expression is $$\sum_{n=0}^{\infty } \frac{t^{n+2} D_V^{n+2} }{n!} \frac{n!}{(n+2)!} = \sum_{n=2}^{\infty } \frac{t^n D_V^n }{n!}$$ exactly as it should ...

9

The paper 'A New Method for Computing Asymptotics of Diagonal Coefficients of Multivariate Generating Functions' by A. Raichev and M. Wilson has the precise machinery that can solve this problem. Get a copy and these brief notes correspond to their symbols, for the diagonal case $$f_{n,n} = [x^n \, y^n ] \frac{I(x)}{J(x)} = [x^n \, y^n ] \frac{(1+x)(1+y)}{... 8 The field of improving convergence of sample averages is known as "enhanced sampling". As Robert pointed out, this is an incredibly hard problem. In my field (theoretical chemistry), we have been struggling with it for the last-half century. The correct way to approach this problem on depends heavily on what information you have available to you. The ... 7 Multilinear equations are hardly easier than general equations. For instance, the multilinear equations$$ \begin{cases} x_0-x_1=0,\\ x_0x_1-x_2=0,\\ x_0x_2-x_3=0,\\ \ldots\\ x_0x_{n-1}-x_n=0 \end{cases} $$simply tell you that x_k=x_0^k for all k=1,\ldots,n. Using this, it is very easy to replace any system of polynomial equations by a system of ... 6 Nemo's representation of F(\eta) in terms of a hypergeometric function can be evaluated without difficulty for large \eta:$$F(\eta)=\frac{\sqrt{3} {\eta}^2 \Gamma \left(\frac{2}{3}\right) \; _1F_2\left(\frac{2}{3};\frac{4}{3},\frac{5}{3};-\frac{{\eta}^3}{27}\right)}{6\pi }-\frac{12 {\eta} \; _1F_2\left(\frac{1}{3};\frac{2}{3},\frac{4}{3};-\frac{{\eta}^3}...

5

It is actually pretty simple if you are comfortable with Taylor series (definitely not MO level, so ask on MSE next time). Let $w=\frac{z-1}{z}$. If $|w|<1$, you are in no trouble computing the expression as it is. So let's consider the case $|w|>1$. Then $\frac1{1-z}=1-\frac 1w$, so your expression in parentheses (the one you really have trouble with) ...

5

There is an interesting though partial answer to your question: Under your assumptions, it is not possible that the signature of $A$ be $(n-1,1)$ (exactly one negative eigenvalue). Proof: With standard notations, $A=D-E-E^T$ where $E$ is strictly triangular and $D$ diagonal. By assumption, $D>0_n$. Notice that the assumption of convergence implies ...

4

We have $f(u) < u$ for $0 < u < 1$ and $f(u) > u$ for $u > 1$, so the fixed point $1$ is unstable. Similarly $f(u) < u$ for $u < -1$ and $f(u) > u$ for $-1 < u < 0$ implies $-1$ is unstable. Since $|f(u)| < |u|$ for $u$ close to $0$, $0$ is stable.

4

For small Hermitian (or real symmetric) matrices, yes, but really this is a hard problem not fully solved. See [1,2] for algorithms. The Cardoso paper  looks at the non-commuting case, but in the commuting case should minimize the off diagonal errors with respect to the Frobeneius norm. I don't know about about getting matrices simultaneously into upper ...

4

See e.g. Y. Karshon, S. Sternberg, and J.Weitsman. The Euler-Maclaurin formula for simple integral polytopes and Y. Karshon, S. Sternberg, and J. Weitsman. Euler-Maclaurin with remainder for a simple integral polytope. For an alternative summation formula -- in terms of integrals only, without using derivatives -- see Approximating sums by integrals only: ...

4

1) No, in general, convergence does not hold because of the presence of spurious poles. A typical result, see , is Theorem. Let $(n_{\nu})_{1}^{\infty}$ be a sequence of positive integers satisfying $n_{\nu}>2 n_{\nu-1}$ for all $\nu$. Then there exists an entire function $$f(z)=\sum_{i=0}^{\infty} a_{i} z^{i}$$ with $a_{i}$ tending to 0 arbitrarily ...

3

Mathematica seems to be plotting the function just fine... If we look a bit at the integrand, it's clear that most of the mass is around $y = \pi/2$ as $x$ increases which should let us introduce a $\sin y$ term and use the antiderative you've already found. We can try to cut the integral at $\pi/2 - 1/x$. Let $$I = \int_0^{\pi/2 - 1/x} 2 e^x \frac{e^{e^x ... 3 Yes, this is true, and there is a proof which closely tracks your intuition. As you know, this estimate can be proved in the continuum by applying the Sobolev embedding twice, first to get \nabla u \in L^p for p<\frac{2d}{d-2}=6, and then once more to get u\in L^\infty. So for simplicity let me discuss how to get discrete versions of the Sobolev ... 2 Turning my previous comments into an answer. What you have is an algebraic Riccati equation: indeed, setting F=-\frac12 I, you get F^TX+XF + B = XAX. Since A and B are positive definite, the pairs (F,A) and (F^T,B) are controllable and observable, so the classical solvability criteria are satisfied and the equation has a unique stabilizing, ... 2 This reference claims to have invented the tabular method as a "novel method": A novel method for calculating the convolution sum of two finite length sequences, J.W. Pierre (1996). Three variations of the tabular method are discussed in The use of spreadsheets to calculate the convolution sum of two finite sequences (2004), citing a 1990 text book. 2 If c_i's are all positive, there are infinitely many such convex C^2 functions. As I have pointed out in my comment above, nonnegativity of c_i's is insufficient to guarantee the existence of a C^2 function. One very simple construction via Bezier curve is as follows. Draw a straight line through each point (x_i,f_i) such that all other points ... 2 Minimum-finding routines (which is what Nelder-Mead/downhill-simplex is) are generally poorly suited to finding zeros of equation systems — if you add the squares of all equations, you get many spurious local minima in addition to the global minimum corresponding to the zero (assuming your equations have a unique zero). If you can't afford to evaluate the ... 1 I understand the question as a request for pointers in the literature to research in the discretization of spacetime. There are two reasons why this is an active research topic, a fundamental and a practical reason: Fundamentally, spacetime might be discrete at the smallest levels (Planck scale); practically, to simulate relativistic quantum field theories (... 1 Find your greatest in absolute value eigenvalue, call it E. If it is positive, the matrix M + 2 EI is positive definite, so do whatever you do for positive definite matrices. If it is negative, M - 2 E I is positive definite. 1 Too long to comment. The square of the distance from a point P to a point on a quadratic Bezier curve (parametrized by t) is a quartic in t. So, choose a Bezier curve and determine this quartic polynomial function. The minima t^* in the range [0,1] can lie at one of the critical points (gradient equal to zero) or the boundaries 0 or 1. The ... 1 Anthony Carapetis has code on his website that runs curve shortening flow. You can find that at http://a.carapetis.com/code and it links to a github account which has the source code. His demonstration is for closed curves though. The issue with open curves is that there is not going to be a unique solution if you don't impose some sort of boundary ... 1 This reference Mulansky, Bernd; Schmidt, Jochen W. Constructive methods in convex C2 interpolation using quartic splines. Numer. Algorithms 12 (1996), no. 1-2, 111–124 may be helpful, but there certainly are more recent ones. 1 This is not a full answer, but I guess it is more than a comment. One way to reduce the task is to apply the spirit of VARPRO to separate the linear coefficients and the non-linear parameters, i.e. alternatingly solve N of the conditions as a linear equation system for the coefficients of the solution in a given N-dimensional subspace, and solve the ... 1 R.B. Paris, An expansion for the sum of a product of an exponential and a Bessel function (2019), equation (2.2):$$\sum_{n=1}^\infty J_0(na)=\frac{1}{a}-\frac{1}{2},\;\;a>0.$$see also The evaluation of single Bessel function sums (2018). Nemo pointed out this should further be restricted to 0<a<2\pi. 1 Here is what I believe is a relevant example: consider the vector field X in \mathbb R^3,$$ X=a_1(x_2, x_3)\frac{\partial}{\partial x_1}+a_2(x_1, x_3)\frac{\partial}{\partial x_2}+a_3(x_1,x_2)\frac{\partial}{\partial x_3}, \quad\text{so that div $X=0$.} $$Assume that a_j\in L^\infty, \frac{\partial a_j}{\partial x_1}, \frac{\partial a_j}{\partial ... 1 Here are the details (for the second definition of Chebyshev nodes). We can define the new continuous function g on the circle by g(z)=f(\frac{z+z^{-1}}2) and consider the interpolation by the trigonometric polynomials Q_n(z)=\sum_{k=0}^n b_k\frac{z^k+z^{-k}}{2} (so that P_n(\frac{z+z^{-1}}{2})=Q_n(z)) on the 2n-th roots of unity. The square of ... 1 Assuming D is bounded, the best estimate that holds almost surely is given by the law of the iterated logarithm https://en.m.wikipedia.org/wiki/Law_of_the_iterated_logarithm 1 The flow X:[0,T]\times \mathbb{R}\to\mathbb{R} defined by X(t,x)= x+t\chi_{\mathbb{R}_+}(x), for all (t,x)\in [0,T]\times \mathbb{R}, is a regular Lagrangian flow solution to (\star) in the sense of Definition (4) of the linked paper (for a.e. initial datum x\in \mathbb{R}, in fact for all, one has (x+t\chi_{\mathbb{R}_+}(x))'=\chi_{\mathbb{R}_+}(... 1 Let H be the Heaviside function (characteristic function of (0,+\infty)). The ODE$$ \dot x=H(x)\text{ on $t>0$}, \quad x(0)=0,  has solutions $x_1(t) = 0$ as well as $x_2(t)=t$. Thus non-uniqueness. Your example is one-dimensional: in that case, Lipschitz continuity of the flux is not required to get uniqueness. Take for instance \$x_0\in \mathbb ...

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